以H3BO3作助熔剂,用高温固相法在1150℃、保温4 h的条件下成功制备了LaMgAl11O19:Mn的单相粉末样品并研究了其真空紫外光激发下的一系列发光特性.在紫外光(254 nm)激发下,LaMgAl11O19:Mn不发光;真空紫外光(147 nm)激发下,观察到Mn2+很强的3d5→3d44s的跃迁发光,峰值位于516 nm,结果表明,Mn2+的掺杂浓度在0.05 mol/mol时发光最强.为了继续增强LaMgAl11O19中Mn2+的发光强度,在固定Mn2+的浓度为0.05 mol/mol的条件下又合成了LaMgAl11019:(Eu2+,Mn2+)与LaMgAl11O19:(Gd3+,Mn2+),利用(Eu2+,Mn2+)和(Gd3+,Mn2+)间存在有效的能量传递的特性,很好的达到了增强Mn2+的发光的目的.
Using H3BO3 as a flux,LaMgAl11O19:Mn2+ and LaMgAl11O19:RE,Mn2+(RE=Eu2+,Gd3+)were synthesized by solid state reaction at 1150℃ for 4 h.The effect of Mn2+ on the luminescent properties excited by vacuum ultraviolet(VUV)was investigated.The strong emission spectrum excited by 147 nm is composed of peaks from Eu2+ and Mn2+ luminescence in LaMgAl11O19:(Eu2+,Mn2+)and only Mn2+ luminescence in LaMgAl11O19:Mn2+ and LaMgAl11O19:(Gd3+,Mn2+).The results show that the green emission of Mn2+ becomes stronger because of efficient energy transferring from Eu2+ and Gd3+ to Mn2+.
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